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CROWDMATH 2024: Where Does the Goldbach Conjecture Hold?

Where Does the Goldbach Conjecture Hold? Polymath project forum
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Where Does the Goldbach Conjecture Hold? Polymath project forum
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Open problem 1
drhong   9
N Nov 19, 2024 by harrypmath
For the case where $r$ is transcendental, we see that $\mathbb{N}[r]\equiv\mathbb{N}[x]$ so no version of goldbach’s conjecture holds. Furthermore, in the case that $r\in\mathbb{N}$, $\mathbb{N}[r]\equiv\mathbb{N}$, so the only known version is Goldbach’s weak conjecture.

Now let $r=\frac{1}{n}$ for some natural number $n$ greater than 1. Then the atoms of $\mathbb{N}[r]$ are of the form $\frac{p}{m}$ where $m$ divides a power of $n$ and $p$ is a prime not dividing $n$. Now take any $\frac{a}{n^k}\in\mathbb{N}[\frac{1}{n}]$. Then by goldbach's weak conjecture, $\frac{a}{n^k}$ can be written as the sum of at most 4 elements of the form $\frac{p}{n^k}$ for primes $p$. For each term, if $p\nmid n$, the term is an atom. Otherwise, it can be written as $\frac{1}{m}$ for some $m\mid n^k$. Let $q$ be a prime greater than $n$. By Dirichlets theorem, there is some $f$ with $1+fq$ prime. Because $1+fq>q>n$, $1+fq\nmid n$. Take some $g$ where $n^g\equiv 1 \pmod q$ and $n^g\ge 1+fq$. Then $n^g=(1+fq)+lq$ for some $l$. So $\frac{1}{m}=\frac{1+fq}{mn^g}+l\frac{q}{mn^g}$ can be wrritten as a sum of $l+1$ atoms. So every element can be written as a sum of at most $4l+4$ atoms.
9 replies
drhong
Sep 22, 2024
harrypmath
Nov 19, 2024
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